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olganol [36]
2 years ago
7

2 1/3 + 3/7 divided by 2 1/3 x 1 3/7​

Mathematics
1 answer:
oee [108]2 years ago
4 0

Answer: \frac{29}{35}

Step-by-step explanation:

1. Make sure all your denominators are the same, you can do this by multiplying them all by 3 or 7, 2 and 1/3 would become 2 7/21, 3/7 would become 9/21, 2 and 1/3, would become 2 and 7/21, and so on and so forth, leaving us with

2\frac{7}{21}+\frac{9}{21} / 2\frac{7}{21} * 1\frac{9}{21}

2. Now, I would go and solve each side of the equation because we divide them, we can add 2\frac{7}{21} and \frac{9}{21}  together to get  2\frac{16}{21}, because 7 + 9 = 16

2\frac{16}{21} / 2\frac{7}{21} * 1\frac{9}{21}

3. We can do the same thing to the other side, by multiplying the two of them. I'm going to convert both of them to just fractions, 2\frac{7}{21} becomes \frac{49}{21} and 1\frac{9}{21} becomes \frac{30}{21}

\frac{49}{21} *  \frac{30}{21}

Now we can reduce them, and for our first fraction divide the top and bottom by 7, giving us  \frac{7}{3} and the second one by 3, which gives us  \frac{10}{7}

\frac{7}{3} * \frac{10}{7} now we cross divide, where we replace the 7's with ones because they're right across from each other to get  \frac{1}{3} and  \frac{10}{1}, or just 10.

\frac{1}{3} * 10 =  \frac{10}{3}

4. So now we can do a similar thing with the first half of our equation and convert it into a fraction and not a mixed number.

\frac{58}{21} / \frac{10}{3}

Now we multiply divide them, and the easiest way I learned how to do this was by keep the first fraction in its place, and flipping the denominator and numerator of the second, and multiplying them.

so   \frac{58}{21} * \frac{3}{10} which if we multiply both sides, you get \frac{174}{210}

I'll save you some times and simplify it for you, the greatest common factor is 6 so we divide the top and bottom by 6 to get

\frac{29}{35}

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<h3>What is a perpendicular bisector of the line segment?</h3>

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To learn more about perpendicular bisector of the line segment from the given link:

brainly.com/question/4428422

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